
Let R be a relation on \[N \times N\]defined by\[(a,b)R(c,d) \Rightarrow ad(b + c) = bc(a + d)\]. R is
A. A partial order relation
B. An equivalence relation
C. An identity relation
D. None of these
Answer
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Hint: We check Reflexivity, symmetricity, and transitivity conditions to know which option is correct. If all three conditions are satisfied then option B is correct, if only two then option A is correct and if the element is related to itself only then option C is correct.
Complete step by step solution: We know from the question relation is \[(a,b)R(c,d) \Rightarrow ad(b + c) = bc(a + d)\]
Now check all the conditions, 1st Reflexive relation,
Let \[a \in a\]and every\[(a,a) \in R\], So \[(c.d) = (a,a)\]
\[(a,b)R(a,b) \Rightarrow ab(b + a) = ba(a + b)\], From this, we can say that R is reflexive.
2nd Symmetric relation, Let\[(a,b) \in R\], So\[(b,a) \in R\]
\[{l}(c,d)R(a,b) \Rightarrow cb(d + a) = da(b + c)\\bc(a + d) = ad(b + c)\]
LHS = RHS
From this reflexivity is confirmed, Now 3rd Transitive relation
If \[(a,b) in R,(b,c) \in R], then (a,c) \in R]
Let [(a,b)R(c,d) \Rightarrow ad(b + c) = bc(a + d)\] ------equation (1)
\[(c,d)R(e,f) \Rightarrow cf(d + e) = de(c + f)\] ---------equation (2)
\[(a,b)R(e,f) \Rightarrow af(b + c) = be(a + f)\] ---------equation (3)
From equation (1) \[\]
\[{l}\dfrac{{b + c}}{{bc}} = \dfrac{{a + d}}{{ad}}\\\dfrac{1}{c} + \dfrac{1}{b} = \dfrac{1}{d} + \dfrac{1}{a}\] ------(a)
From equation (2)\[\]
\[{l}\dfrac{{d + e}}{{de}} = \dfrac{{c + f}}{{cf}}\\\dfrac{1}{e} + \dfrac{1}{d} = \dfrac{1}{f} + \dfrac{1}{c}\] -------(b)
Adding (a) and (b), we get
\[{l}\dfrac{1}{b} + \dfrac{1}{c} + \dfrac{1}{d} + \dfrac{1}{e} = \dfrac{1}{a} + \dfrac{1}{c} + \dfrac{1}{d} + \dfrac{1}{f}\\\dfrac{1}{b} + \dfrac{1}{e} = \dfrac{1}{a} + \dfrac{1}{f}\\\dfrac{{b + e}}{{be}} = \dfrac{{a + f}}{{af}}\\af(b + e) = be(a + f)\]
We get equation (3) from (1) and (2), So we say that R is transitive, from checking the above three conditions we say that R is an equivalence relation.
Thus, Option (B) is correct.
Note: There is a little difference between reflexivity and identity relation. In reflexive relation, every element of a set is related to itself but in identity, every element of a set is related to itself only.
Complete step by step solution: We know from the question relation is \[(a,b)R(c,d) \Rightarrow ad(b + c) = bc(a + d)\]
Now check all the conditions, 1st Reflexive relation,
Let \[a \in a\]and every\[(a,a) \in R\], So \[(c.d) = (a,a)\]
\[(a,b)R(a,b) \Rightarrow ab(b + a) = ba(a + b)\], From this, we can say that R is reflexive.
2nd Symmetric relation, Let\[(a,b) \in R\], So\[(b,a) \in R\]
\[{l}(c,d)R(a,b) \Rightarrow cb(d + a) = da(b + c)\\bc(a + d) = ad(b + c)\]
LHS = RHS
From this reflexivity is confirmed, Now 3rd Transitive relation
If \[(a,b) in R,(b,c) \in R], then (a,c) \in R]
Let [(a,b)R(c,d) \Rightarrow ad(b + c) = bc(a + d)\] ------equation (1)
\[(c,d)R(e,f) \Rightarrow cf(d + e) = de(c + f)\] ---------equation (2)
\[(a,b)R(e,f) \Rightarrow af(b + c) = be(a + f)\] ---------equation (3)
From equation (1) \[\]
\[{l}\dfrac{{b + c}}{{bc}} = \dfrac{{a + d}}{{ad}}\\\dfrac{1}{c} + \dfrac{1}{b} = \dfrac{1}{d} + \dfrac{1}{a}\] ------(a)
From equation (2)\[\]
\[{l}\dfrac{{d + e}}{{de}} = \dfrac{{c + f}}{{cf}}\\\dfrac{1}{e} + \dfrac{1}{d} = \dfrac{1}{f} + \dfrac{1}{c}\] -------(b)
Adding (a) and (b), we get
\[{l}\dfrac{1}{b} + \dfrac{1}{c} + \dfrac{1}{d} + \dfrac{1}{e} = \dfrac{1}{a} + \dfrac{1}{c} + \dfrac{1}{d} + \dfrac{1}{f}\\\dfrac{1}{b} + \dfrac{1}{e} = \dfrac{1}{a} + \dfrac{1}{f}\\\dfrac{{b + e}}{{be}} = \dfrac{{a + f}}{{af}}\\af(b + e) = be(a + f)\]
We get equation (3) from (1) and (2), So we say that R is transitive, from checking the above three conditions we say that R is an equivalence relation.
Thus, Option (B) is correct.
Note: There is a little difference between reflexivity and identity relation. In reflexive relation, every element of a set is related to itself but in identity, every element of a set is related to itself only.
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